Find the integrand and the limits of integration in terms of u

Find the integrand and the limits of integration in terms of u
Let u=x^3+3

∫[0,1,(3x^2√(x^3+3)),] a=0 b=1

With u = x^3 + 3, du = 3x^2 dx. With x = 0, u = 3 and with x = 1, u = 1^3 + 3 = 4.
Thus the integral becomes

integral[(sqrt(u)) du] (from u=3 to 4) = u^(3/2)/(3/2) (from u = 3 to 4) =

(2/3) * (4^(3/2) - 3^(3/2)) = (2/3) * (8 - 3sqrt(3)) = 1.869 to 3 decimal places.